This is really correct: the stabilization isomorphism isn't compatible with the multiplication. This is a problem even in ordinary homological algebra and is largely a consequence of pretending that integer indices are sufficient. Let me try to illustrate.

Suppose that $K$ is a homotopy-commutative differential graded algebra and that $X$ and $Y$ are chain complexes. For any integer $i$ we can define $$ K^{-i}(X) = [\Bbb Z[i] \otimes X, K] $$ by chain homotopy classes of maps. This gives an external product $$ K^{-i}(X) \otimes K^{-j}(Y) \to [\Bbb Z[i] \otimes X \otimes \Bbb Z[j] \otimes Y, K \otimes K] \to [\Bbb Z[i] \otimes \Bbb Z[j] \otimes X \otimes Y, K] $$ using the product of $K$. This points out that we need to choose an identification of $\Bbb Z[i] \otimes \Bbb Z[j]$ with $\Bbb Z[i+j]$ to proceed further. Choosing one makes it land in $K^{-(i+j)}(X \otimes Y)$.

The reason why I'm going to this length is that, when you describe the exterior product stability isomorphism $h^\ast X \to h^{\ast + 1}(S^1 \wedge X)$, it corresponds in homological algebra to choosing an element $$ \sigma \in K^1(\Bbb Z[1]) = [\Bbb Z[-1] \otimes \Bbb Z[1], K] \cong [\Bbb Z, K] = H_0 K $$ lifting the unit. Again, there is this last casual identification of $\Bbb Z[-1] \otimes \Bbb Z[1]$ with $\Bbb Z[0]$. Without making casual identifications early, the exterior-product stabilization isomorphism you describe is most naturally the exterior product map $$ \begin{align*} K^{-i}(X) &= [\Bbb Z[i] \otimes X, K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[1] \otimes \Bbb Z[i] \otimes X, K \otimes K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[i] \otimes \Bbb Z[1] \otimes X, K] \\ &\cong K^{-(-1 +i)}(\Bbb Z[1] \otimes X). \end{align*} $$ Note in particular the swap of the factors $\Bbb Z[1]$ and $\Bbb Z[i]$ at the second-last step. This introduces a sign of $(-1)^i$ into the standard identifications that we should probably account for in the suspension isomorphism. If you repeat this $i$ times to land in $K^0$, you introduce a sign of $(-1)^{\binom{i+1}{2}}$, and so the net sign discrepancy introduced by this on the source and target has parity $$ \binom{i+1}{2} + \binom{j+1}{2} - \binom{i+j+1}{2} \equiv ij \mod 2. $$

In homotopy theory we have the same issue except that we are permuting sphere factors, and have less canonical justification for suspending vs desuspending. Adams wrote a rather scathing criticism of these types of casual identification in section 6 of "Prerequisites for Carlsson's lecture".

This is really correct: the stabilization isomorphism isn't immediately compatible with the multiplication. This is a problem even in ordinary homological algebra and is largely a consequence of pretending that integer indices are sufficient. Let me try to illustrate.

Suppose that $K$ is a homotopy-commutative differential graded algebra and that $X$ and $Y$ are chain complexes. For any integer $i$ we can define $$ K^{-i}(X) = [\Bbb Z[i] \otimes X, K] $$ by chain homotopy classes of maps. This gives an external product $$ K^{-i}(X) \otimes K^{-j}(Y) \to [\Bbb Z[i] \otimes X \otimes \Bbb Z[j] \otimes Y, K \otimes K] \to [\Bbb Z[i] \otimes \Bbb Z[j] \otimes X \otimes Y, K] $$ using the product of $K$. This points out that we need to choose an identification of $\Bbb Z[i] \otimes \Bbb Z[j]$ with $\Bbb Z[i+j]$ to proceed further. Choosing one makes it land in $K^{-(i+j)}(X \otimes Y)$.

The reason why I'm going to this length is that, when you describe the exterior product stability isomorphism $h^\ast X \to h^{\ast + 1}(S^1 \wedge X)$, it corresponds in homological algebra to choosing an element $$ \sigma \in K^1(\Bbb Z[1]) = [\Bbb Z[-1] \otimes \Bbb Z[1], K] \cong [\Bbb Z, K] = H_0 K $$ lifting the unit. Again, there is this last casual identification of $\Bbb Z[-1] \otimes \Bbb Z[1]$ with $\Bbb Z[0]$. Without making casual identifications early, the exterior-product stabilization isomorphism you describe is most naturally the exterior product map $$ \begin{align*} K^{-i}(X) &= [\Bbb Z[i] \otimes X, K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[1] \otimes \Bbb Z[i] \otimes X, K \otimes K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[i] \otimes \Bbb Z[1] \otimes X, K] \\ &\cong K^{-(-1 +i)}(\Bbb Z[1] \otimes X). \end{align*} $$ Note in particular the swap of the factors $\Bbb Z[1]$ and $\Bbb Z[i]$ at the second-last step. This introduces a sign of $(-1)^i$ into the standard identifications that we should probably account for in the suspension isomorphism. If you repeat this $i$ times to land in $K^0$, you introduce a sign of $(-1)^{\binom{i+1}{2}}$, and so the net sign discrepancy introduced by this on the source and target has parity $$ \binom{i+1}{2} + \binom{j+1}{2} - \binom{i+j+1}{2} \equiv ij \mod 2. $$

In homotopy theory we have the same issue except that we are permuting sphere factors, and have less canonical justification for suspending vs desuspending. Adams wrote a rather scathing criticism of these types of casual identification in section 6 of "Prerequisites for Carlsson's lecture". Schwede's book on symmetric spectra also discusses systematically keeps track of the order of indices in terms like $i+j$ to make these types of issues more transparent.

This is really correct: the stabilization isomorphism isn't compatible with the multiplication. This is a problem even in ordinary homological algebra and is largely a consequence of pretending that integer-indices are sufficient. Let me try to illustrate.

Suppose that $K$ is a homotopy-commutative differential graded algebra and that $X$ and $Y$ are chain complexes. For any integer $i$ we can define $$ K^{-i}(X) = [\Bbb Z[i] \otimes X, K] $$ by chain homotopy classes of maps. This gives an external product $$ K^{-i}(X) \otimes K^{-j}(Y) \to [\Bbb Z[i] \otimes X \otimes \Bbb Z[j] \otimes Y, K \otimes K] \to [\Bbb Z[i] \otimes \Bbb Z[j] \otimes X \otimes Y, K] $$ using the product of $K$. This points out that we need to choose an identification of $\Bbb Z[i] \otimes \Bbb Z[j]$ with $\Bbb Z[i+j]$ to proceed further. Choosing one makes it land in $K^{-(i+j)}(X \otimes Y)$.

The reason why I'm going to this length is that, when you describe the exterior product stability isomorphism $h^\ast X \to h^{\ast + 1}(S^1 \wedge X)$, it corresponds in homological algebra to choosing an element $$ \sigma \in K^1(\Bbb Z[1]) = [\Bbb Z[-1] \otimes \Bbb Z[1], K] \cong [\Bbb Z, K] = H_0 K $$ lifting the unit. Again, there is this last casual identification of $\Bbb Z[-1] \otimes \Bbb Z[1]$ with $\Bbb Z[0]$. Without making casual identifications early, the exterior-product stabilization isomorphism you describe is most naturally the exterior product map $$ \begin{align*} K^{-i}(X) &= [\Bbb Z[i] \otimes X, K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[1] \otimes \Bbb Z[i] \otimes X, K \otimes K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[i] \otimes \Bbb Z[1] \otimes X, K] \\ &\cong K^{-(-1 +i)}(\Bbb Z[1] \otimes X). \end{align*} $$ Note in particular the swap of the factors $\Bbb Z[1]$ and $\Bbb Z[i]$ at the second-last step, which introduces a sign of $(-1)^i$ into the standard identifications. If you repeat this $i$ times to land in $K^0$, you introduce a sign of $(-1)^{\binom{i+1}{2}}$, and so the net sign discrepancy introduced by this on the source and target has parity $$ \binom{i+1}{2} + \binom{j+1}{2} - \binom{i+j+1}{2} \equiv ij \mod 2. $$

In homotopy theory we have the same issue except that we are permuting sphere factors, and have less canonical justification for suspending vs desuspending.

This is really correct: the stabilization isomorphism isn't compatible with the multiplication. This is a problem even in ordinary homological algebra and is largely a consequence of pretending that integer indices are sufficient. Let me try to illustrate.

Suppose that $K$ is a homotopy-commutative differential graded algebra and that $X$ and $Y$ are chain complexes. For any integer $i$ we can define $$ K^{-i}(X) = [\Bbb Z[i] \otimes X, K] $$ by chain homotopy classes of maps. This gives an external product $$ K^{-i}(X) \otimes K^{-j}(Y) \to [\Bbb Z[i] \otimes X \otimes \Bbb Z[j] \otimes Y, K \otimes K] \to [\Bbb Z[i] \otimes \Bbb Z[j] \otimes X \otimes Y, K] $$ using the product of $K$. This points out that we need to choose an identification of $\Bbb Z[i] \otimes \Bbb Z[j]$ with $\Bbb Z[i+j]$ to proceed further. Choosing one makes it land in $K^{-(i+j)}(X \otimes Y)$.

The reason why I'm going to this length is that, when you describe the exterior product stability isomorphism $h^\ast X \to h^{\ast + 1}(S^1 \wedge X)$, it corresponds in homological algebra to choosing an element $$ \sigma \in K^1(\Bbb Z[1]) = [\Bbb Z[-1] \otimes \Bbb Z[1], K] \cong [\Bbb Z, K] = H_0 K $$ lifting the unit. Again, there is this last casual identification of $\Bbb Z[-1] \otimes \Bbb Z[1]$ with $\Bbb Z[0]$. Without making casual identifications early, the exterior-product stabilization isomorphism you describe is most naturally the exterior product map $$ \begin{align*} K^{-i}(X) &= [\Bbb Z[i] \otimes X, K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[1] \otimes \Bbb Z[i] \otimes X, K \otimes K] \\ &\to [\Bbb Z[-1] \otimes \Bbb Z[i] \otimes \Bbb Z[1] \otimes X, K] \\ &\cong K^{-(-1 +i)}(\Bbb Z[1] \otimes X). \end{align*} $$ Note in particular the swap of the factors $\Bbb Z[1]$ and $\Bbb Z[i]$ at the second-last step. This introduces a sign of $(-1)^i$ into the standard identifications that we should probably account for in the suspension isomorphism. If you repeat this $i$ times to land in $K^0$, you introduce a sign of $(-1)^{\binom{i+1}{2}}$, and so the net sign discrepancy introduced by this on the source and target has parity $$ \binom{i+1}{2} + \binom{j+1}{2} - \binom{i+j+1}{2} \equiv ij \mod 2. $$

In homotopy theory we have the same issue except that we are permuting sphere factors, and have less canonical justification for suspending vs desuspending. Adams wrote a rather scathing criticism of these types of casual identification in section 6 of "Prerequisites for Carlsson's lecture".